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3 months ago

Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

1 answer:

7 0

The de Broglie wavelength can be defined as $\lambda = \frac{h}{mv}$ , where h stands for Planck's constant, m denotes mass, and v represents velocity.

$h = 6.63*10^{-34}$

The mass for a hydrogen atom is $m = 1.67*10^{-27}kg$

With v equal to 440 m/s

Plugging in the values

Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So, \lambda =902 pm$

This means the de Broglie wavelength (measured in picometers) for a hydrogen atom moving at 440 m/s is 902 pm

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